Certainly! To solve this problem, let's start by defining variables, generate a system of equations, and solve them to determine how many adult and children tickets were sold.
Step 1: Define Variables
Let's define [tex]\( x \)[/tex] as the number of adult tickets sold and [tex]\( y \)[/tex] as the number of children tickets sold.
Step 2: Set Up the System of Equations
We have two pieces of information that will lead to two equations.
1. The total number of tickets sold is 350.
2. The total amount of money earned from selling these tickets is [tex]$4,410.
The first piece of information can be translated into an equation:
\[ x + y = 350 \]
The second piece of information can be represented as:
\[ 15x + 8y = 4410 \]
where $[/tex]15 is the price of each adult ticket and [tex]$8 is the price of each child ticket.
Step 3: Solve the System of Equations
We now have two equations with two variables:
\[
\begin{align*}
x + y & = 350 \\
15x + 8y & = 4410
\end{align*}
\]
Using substitution or elimination method, we can solve for \( x \) and \( y \). Let's use the elimination method in this case:
To eliminate \( y \), we can multiply the first equation by 8 to make the coefficient of \( y \) in this modified equation equal to the coefficient in the second equation.
Multiplying the entire first equation by 8 gives us:
\[ 8x + 8y = 2800 \]
Now subtract this new equation from the second equation:
\[
\begin{align*}
(15x + 8y) - (8x + 8y) & = 4410 - 2800 \\
15x - 8x & = 1610 \\
7x & = 1610 \\
x & = \frac{1610}{7} \\
x & = 230
\end{align*}
\]
Now that we have the value of \( x \), we can substitute it back into the first equation to solve for \( y \):
\[
\begin{align*}
230 + y & = 350 \\
y & = 350 - 230 \\
y & = 120
\end{align*}
\]
Step 4: Answer the Question
We have determined that the theater sold 230 adult tickets and 120 children tickets.
So, the theater sold 230 adult tickets at $[/tex]15 each, and 120 children tickets at $8 each.
